CONTINUITY & DIFFERENTIABILITY

EXERCISE 1(A)

1. (d)

L.H.L. at x = 3, x 3 x 3lim f (x) lim (x )

=

h 0lim(3 h )

=3 …..(i)

R.H.L. at x = 3, x 3 x 3lim f (x) lim(3x 5)

=

h 0lim{3(3 h) 5}

= 4 …..(ii)

Value of function 4)3( f …..(iii)

For continuity at x = 3

Limit of function = value of function 3 + = 4 = 1.

2. (c)

If function is continuous at x = 0, then by the definition of continuity x 0

f (0) limf (x)

Since f(0) = k. Hence, x 0

1f (0) k lim(x) sin

x

k = 0 (a finite quantity lies between –1 to 1)

k = 0.

3. (c)

Since f(x) is continuous at x = 1,

x 1 x 1lim f (x) lim f (x) f (1)

…..(i)

Now h 0x 1

lim f (x) limf (1 h)

= h 0lim2(1 h) 1 3

i.e., x 1lim f (x) 3

Similarly, h 0x 1

lim f (x) limf (1 h)

=h 0lim5(1 h) 2

i.e., x 1lim f (x) 3

So according to equation (i), we have k = 3.

4. (d)

We have x 0 x 0

1limf (x) limsin

x = An oscillating number which oscillates between –1 and 1.

Hence, x 0limf (x)

does not exist.

Consequently f(x) cannot be continuous at x = 0 for any value of k.

5. (c)

LHL = 2

h 0x 1lim f (x) limm(1 h) m

RHL = h 0x 1

lim f (x) lim2(1 h) 2

and f (1) m

Function is continuous at x = 1, LHL = RHL = f(1)

Therefore m = 2.

6. (a)

1/x

x 0 x 0

1lim(cos x) k lim log(cos x) log k

x

x 0 x 0

1lim limlogcos x log k

x

ex 0

1lim 0 log k k 1

x .

7. (b)

Since f is continuous at x4

;

h 0 h 0f f h f h

4 4 4

2b a

4 4

2b a

Also as f is continuous at x2

;

h 0 h 0

f limf h limf h2 2 2

2b a b a b .

Hence (–1, 1) & (0, 0) satisfy the above relations.

8. (c)

h 0x 1

lim f (x) limf (1 h)

= h 0lim 2 sin (1 h)

2

= 3

Similarly, h 0x 1

lim f (x) limf (1 h)

= h 0lima(1 h) b

= a + b

f(x) is continuous at x = 1 so x 1 x 1lim f (x) lim f (x) f (1)

a b 3 ..…..(i)

Again,h 0x 2

lim f (x) limf (2 h)

=h 0lima(2 h) b

= 2a b

and h 0x 2

lim f (x) limf (2 h)

= h 0lim tan (2 h)

8

= 1

f(x) is continuous in ( ,6) , so it is continuous at x 2 also, so

x 2 x 2lim f (x) lim f (x) f (2)

2a b 1 ..….(ii)

Solving (i) and (ii) a = - 2, b = 5.

9. (a)

x x

2 2

lim f (x) , lim f (x)2 2

Since x x

2 2

lim f (x) lim f (x)

,

Function is discontinuous at x2

10. (b)

2

2x 0 x 0

2sin 3xlim f (x) lim 3 6

(3x)

and

x 0 x 0 x 0

xlim f (x) lim lim 9 x 3 6

9 x 3

Hence a 6 .

11. (c)

The function 2

1f x

x x 6

is discontinuous at 2 points.

The function 2

1f x

x x 6

&

1g x

x 1

2

1g f x

x x 7

g(f(x)) is discontinuous at 4 points.

Hence, the composite f g x is discontinuous at three points 2 3

x ,1 &3 2

12. (b)

x 0 x 0

ln b ln(a x) ln a ln a ln a ln(b x) ln bln bln(a x) ln a ln(b x)lim lim

x x

x 0 x 0

ln(a x) ln a ln(b x) ln bln blim ln a lim

x x

x 0 x 0

x xln 1 ln 1

ln b ln aa blim lim

x xa b

a b

b aln b aln b ln a

a b ab

13. (b)

2

2x 2 x 2

x 5x 6 (x 3) 1f (2) 2, f (2 ) lim lim

x 4 (x 2) 4

14. (c)

Clearly from curve drawn of the given function )(xf , it is discontinuous at 0x .

15. (b)

a

3|tan x|

tan 6x

tan3x

(1 | tan x |) , x 06

f (x) b , x 0

e , 0 x6

For f(x) to be continuous at x = 0

(0,–1)

O

(0,1/4)

y

y = x2, x > 0

x

y = x-1, x < 0

y'

x'

x 0 x 0lim f (x) f (0) lim f (x)

x 0

aalim 1 |tan x| 1

3|tan x| a /33|tan x|

x 0lim(1 | tan x |) e e

Now,

tan6x tan3xtan6x .6x .3x26x 3xtan3x

x 0 x 0lim e lim e e

a/3 2e b e a 6 and 2b e .

16. (d)

Let x

f x ln4

x 4 x 4

xlim xf x lim x ln 0

4

17. (a)

Note that [x 2] 0 if 0 x 2 1

i.e. [x 2] 0 if 2 x 1.

Thus domain of f is R [ 2, 1)

We have sin[x 2]

is continuous at all points of R [ 2, 1) and [x] is continuous on

R I, where I denotes the set of integers.

Thus the points where f can possibly be discontinuous are……, ...........,2,10,1,2,3 But for

1 x 0,[x 1] 0 and sin[x 2]

is defined.

Therefore f(x) = 0 for 1 x 0.

Also )(xf is not defined on 2 x 1.

Hence set of points of discontinuities of f(x) is I { 1}.

18. (b)

1

1x 0

2x sin xf (x) lim f (0)

2x tan x

,

0form

0

Applying L-Hospital’s rule, 2

x 0

2

12

1 xf (0) lim

12

1 x

2 1 1

2 1 3

19. (d)

For continuity at all x R, we must have

x ( /2)

f lim (4sin x)2

x ( /2)

lim (a sin x b)

4 a b …..(i)

and x ( /2)

f lim (a sin x b) a b2

x ( /2)lim (cos x) 0

0 a b ….(ii)

From (i) and (ii), a 2 and b 2 .

20. (a)

x 5

f (5) limf (x)

2

2x 5

x 10x 25lim

x 7x 10

2

x 5

(x 5) 5 5lim 0

(x 2)(x 5) 5 2

.

21. (c)

For continuity at 0, we must have x 0

f (0) limf (x)

cot x

x 0lim(x 1)

xcot x1

x

x 0lim (1 x)

x 0

xlim

1 tan xx

x 0lim (1 x)

= e.

22. (a)

Conceptual question

23. (c)

f(x) is continuous at x3

, then

x /3lim f (x) f (0)

or

x /3

3x1 sin

2lim3x

,0

form0

Applying L-Hospital’s rule, x /3

3 3xcos

2 2lim 03

24. (d)

If f(x) is continuous at x = 0 then,

x 0

f (0) limf (x)

x 0

2 x 4lim

sin 2x

,

0 form

0

Using L–Hospital’s rule, x 0

1

12 x 4f (0) lim

2cos 2x 8

.

25. (d)

2x 2 3x x 1,2

F(x) will be continuous only at x = 1 & 2.

26. (b)

11

2 2 xf (x) x e

and f (2) k

If f(x) is continuous from right at x = 2 then x 2lim f (x) f (2) k

11

2 2 x

x 2lim x e k

h 0k limf (2 h)

11

2 2 (2 h)

h 0k lim (2 h) e

1

2 1/h

h 0k lim 4 h 4h e

1k [4 0 0 e ]

1k

4 .

27. (c)

2

x x 2

x x x2cos 2sin cos

2 2 2limf (x) limx x x

2cos 2sin cos2 2 2

x x

x xcos sin

x2 2lim lim tanx x 4 2

cos sin2 2

At x , f ( ) tan 14

.

28. (c)

L.H.L. 1 1x 0 x 0

4 kx 4 kx 2kx 1 klim lim

x x 24 kx 4 kx

R.H.L. 2

x 0 x 0

2x 3x xlim lim 2x 3 3

sin x sin x

Since it is continuous, hence L.H.L = R.H.L k 6 .

29. (c)

|| x is continuous at 0x and x

x || is discontinuous at 0x

x

xxxf

||||)( is discontinuous at 0x .

30. (b)

x x

x 0 x 0

x e 1 x e 1lim lim 0

| tan x | tan x

x x

x 0 x 0

x e 1 x e 1lim lim 0

| tan x | tan x

So f(x) is continuous at x = 0.

Now

x

x

x 0 x 0

x e 10

x e 1| tan x |L.H.D. lim lim 1

x 0 tan x x

x

x

x 0 x 0

x e 10

x e 1| tan x |R.H.D. lim lim 1

x 0 tan x x

L.H.D. R.H.D.

F(x) is continuous but not differentiable at x = 0

31. (a)

We have,

x, x 0

1 xx

f (x) 0 , x 01 | x |

x, x 0

1 x

;

h 0

f ( h) f (0)L.H.D. lim

h

=

h 0

h0

1 hlimh

= 1

h 0

f (h) f (0)R.H.D. lim

h

=

h 0

hlim 0

1 h

h

= h 0

1lim

1 h = 1

So, f(x) is differentiable at x = 0; Also f(x) is differentiable at all other points.

Hence, f(x) is everywhere differentiable.

32. (b)

Let f (x) | x 1| | x 3| =

(x 1) (x 3) , x 1

(x 1) (x 3) , 1 x 3

(x 1) (x 3) , x 3

=

2x 4 , x 1

2 , 1 x 3

2x 4 , x 3

Since, f (x) = 2 for 1 x 3 . Therefore f '(x) 0 for all x (1,3) .

Hence, f '(x) 0 at x 2 .

33. (a)

We have, 2 2

2x 0 x 0 x 0

sin x sin xlimf (x) lim lim x 1 0 0 f (0)

x x

So, f(x) is continuous at x = 0,

f (x) is also derivable at

x = 0, because 2

2x 0 x 0

f (x) f (0) sin xlim lim

x 0 x

= 1

exists finitely.

34. (b)

It is evident from the graph of f (x) = | log | x || than

f (x) is everywhere continuous but not differentiable

at x 1.

35. (a)

f (x) [x]sin( x)

If x is just less than k, [x] = k – 1. f (x) (k 1)sin( x) , when x k k I

Now L.H.D. at x k ,

= x k

(k 1)sin( x) ksin( k)lim

x k

=

x k

(k 1)sin( x)lim

(x k)

[as sin( k) 0 k I]

Y

Y'

X X' (1, 0) (–1, 0)

O

f(x)=|log|x||

= h 0

(k 1)sin( (k h))lim

h

[Let x (k h) ]

= k 1

h 0

(k 1)( 1) sin hlim

h

= k 1

h 0

sin hlim(k 1)( 1) ( )

h

= k(k 1)( 1) = k( 1) (k 1) .

36. (a)

We have,

2x 1, x 0

f (x) | x | | x 1| 1 , 0 x 1

2x 1, x 1

Since, x 1 x 1 x 1 x 1lim f (x) lim1 1, lim f (x) lim(2x 1) 1

and f (1) 2 1 1 1

x 1 x 1lim f (x) lim f (x) f (1)

. So, f(x) is continuous at x = 1.

Now, h 0 h 0x 1

f (x) f (1) f (1 h) f (1) 1 1lim lim lim 0

x 1 h h

and

h 0x 1

f (x) f (1) f (1 h) f (1)lim lim

x 1 h

=

h 0

2(1 h) 1 1lim 2

h

.

(LHD at x = 1) (RHD atx = 1).

So, f(x) is not differentiable at x = 1.

Alternately

By graph, it is clear that the function is not differentiable at x = 0, 1 as there it has sharp edges.

37. (c)

Here f (x) | x 1| | x 1|

2x , when x 1

f (x) 2 , when 1 x 1

2x , when x 1

Alternate

The graph of the function is shown alongside,

From the graph it is clear that the function is continuous at all real

x, also differentiable at all real x except at ;1x Since sharp

edges at 1x and 1x .

At 1x we see that the slope from the right i.e., R.H.D. = 2,

while slope from the left i.e., L.H.D.= 0

Similarly, at 1x it is clear that R.H.D. = 0 while L.H.D. = – 2

Here,

2 , x 1

f '(x) 0 , 1 x 1

2 , x 1

(No equality on –1 and +1)

X x =1/2

y = –2x+1 y=1

x =1

y=2x–1

X’ –1 O 1

X

Y

2

Y’

Now, at x 1, f '(1 ) 2 while f '(1 ) 0 and

at x 1, f '( 1 ) 0 while f '( 1 ) 2

Thus, f(x) is not differentiable at x 1 .

38. (d)

1

2

x 1 x 1lim f x lim ax bx 2 a b 2

and

2

x 1 x 1lim f x lim bx ax 4 b a 4

For continuity a b 2 b a 4 a b 1... i

Now 2ax b , x 1

f '(x) R.H.D. 2a b & L.H.D. 2b a2bx a , x 1

For differentiability 2a b 2b a a b... ii

From (i) & (ii) no value of (a, b) is possible.

39. (b)

3 3 3 3 2 2f (x) g(x) x f (x) g(x) x

h(x) e h ' x e 3 f x f ' x 3 g x g ' x 1

2 2g x f (x)h ' x h x 3 f x 3 g x 1

f x g x

x ch ' x h x h x e

Now 6 x 1h 5 e h x e

11Hence h 10 e

40. (c)

[2 h] 2,[2 h] 1,[1 h] 1,[1 h] 0

At x = 2, we will check RHL LHL f 2

h 0

RHL lim | 4 2h 3|[2 h] 2,f 2 1.2 2

h 0

LHL lim | 4 2h 3|[2 h] 1,R L

, not continuous

At x 1,RHL lim | 2 2h 3|[1 h] 1.1 1,

f 1 | 1| [1] 1

h 0

LHL limsin (1 h) 12

continuous at x = 1

R.H.D.h 0

| 2 2h 3 | [1 h] 1lim

h

h 0 h 0

| 1|.1 1 1 1lim lim 0

h h

L.H.D. h 0

| 2 2h 3 | [1 h] 1lim

h

h 0 h 0

1.0 1 1lim lim

h h

Since R.H.D. L.H.D. not differentiable. at x = 1.

41. (b)

Clearly, f (x) is differentiable for all non-zero values of x,

For x 0 , we have

2

2

x

x

xef '(x)

1 e

Now, (L.H.D. at x = 0)

= h 0x 0

f (x) f (0) f (0 h) f (0)lim lim

x 0 h

=

2 2h h

h 0 h 0

1 e 1 elim lim

h h

=2

h2

2h 0 h

e 1 1lim 1

h e

and, (RHD at x = 0) =

2h

h 0x 0

f (x) f (0) 1 e 0lim lim

x 0 h

=

2

2

h

2h 0 h

e 1 1lim 1

h e

.

So, )(xf is not differentiable at x 0 ,

Hence, the points of differentiability of f (x) are ( ,0) (0, )

42. (a)

We have,

sin x

sin x

e , x 02

f (x)

e ,0 x2

Clearly, )(xf is continuous and differentiable for all non-zero x.

Now, sin x

x 0x 0lim f (x) lime

= 1 and sin x

x 0x 0lim f (x) lime 1

Also, 0f (0) e 1

So, f(x) is continuous for all x.

(LHD at x = 0) = x x 0

x 0

x 0

d(e ) (e ) e 1

dx

(RHD at 0x ) = x x

x 0

x 0

d(e ) ( e ) 1

dx

So, f(x) is not differentiable at x = 0.

43. (b)

We have, 2f (x) 1 1 x . The domain of definition of f (x) is [–1, 1].

For x 0,x 1, x 1 we have 22

1 xf '(x)

1 x1 1 x

Since f (x) is not defined on the right side of x 1 and on the left side of x 1 .

Also, f '(x) when x 1 or x 1 .

So, we check the differentiability at x = 0.

Now, (LHD at x 0 ) = x 0 h 0

f (x) f (0) f (0 h) f (0)lim lim

x 0 h

= 2 42

h 0 h 0

1 {1 (1/ 2)h (3 / 8)h ....}1 1 h 0lim lim

h h

= 2

h 0

1 3 1lim h .....

2 8 2

Similarly, (RHD at x = 0) = 2

1

Hence, f (x) is not differentiable at x = 0.

44. (d) Since )(xf is differentiable at ,cx therefore it is continuous at cx .

Hence, )()(lim cfxfcx

.

45. (c)

2(x 3x 2) (x 1)(x 2) 0 When 1x or ,2

And 2(x 3x 2) (x 1)(x 2) 0 when 21 x

Also cos | x | cos x

2 2f (x) (x 4)(x 3x 2) cos x, 1 x 2

and 2 2f (x) (x 4)(x 3x 2) cos x, x 1 or x 2

Evidently )(xf is not differentiable at x = 1.

46. (b)

f (0) 0 and

1 1

|x| x2f (x) x e

R.H.L. = 2

2/h

2/hh 0 h 0

hlim(0 h) e lim 0

e

L.H.L. =

1 1

2 h h

h 0lim(0 h) e 0

f(x) is continuous at x = 0.

h 0

f (0 h) f (0)R.H.D. at (x 0) lim

h

=

2 2/h2/h

h 0

h elim he 0

h

h 0

f (0 h) f (0)L.H.D. at (x 0) lim

h

=

1 1

2 h h

h 0 h 0

h elim lim h 0

h

F(x) is differentiable at x = 0

47. (d)

3 2

x 0

1limf (x) x sin 0

x

as 2 1

0 sin 1x

and x 0

Therefore f (x) is continuous at x 0 .

Also, the function 3 2 1

f (x) x sinx

is differentiable because

3 2

2 2

h 0 h 0

1h sin 0

1hRHD lim lim h sin 0h h

,

3

h 0

1h sin

hLHD lim 0

h

.

48. (b)

49. (d)

50. (c)

h 0lim1 (2 h) 3

,

h 0lim 5 (2 h) 3

, f (2) 3

Hence, f is continuous at x 2

Now h 0

5 (2 h) 3RHD lim 1

h

h 0

1 (2 h) 3LHD lim 1

h

f(x) is not differentiable at x 2 .

51. (c)

g(x) | f (| x |) | 0 . So )(xg cannot be onto.

If f (x) is one-one and 1 2f (x ) f (x ) then 1 2g(x ) g(x ) .

So, ‘ )(xf is one-one’ does not ensure that )(xg is one-one.

If f(x) is continuous for x R, | f (| x |) | is also continuous for x R .

So the answer (c) is correct.

The fourth answer (d) is not correct as f(x) being differentiable does not ensure |f(x)| being

differentiable.

52. (b)

Given f (4) 6,f (4) 1

x 4 x 4

xf (4) 4f (x) xf (4) 4f (4) 4f (4) 4f (x)lim lim

x 4 x 4

x 4 x 2

(x 4)f (4) f (x) f (4)lim 4lim

x 4 x 4

= f (4) 2f ' 4 4

53. (c)

f x 2y 2f x f y 2f ' x 2y 2f x f ' y {partially differentiating w.r.to y}

For x = 5 & y = 0, f ' 5 f 5 f ' 0 f ' 5 6

54. (c)

By L’hospital’s rule

2 22 2 2 2

2x 2 x 2

g(x)g ' x f (2) f (x)f ' x g (2)g (x)f (2) f (x)g (2)lim lim

x 4 x

1 4 9 3 2 115

2

55. (b)

Given 2

5f (2x) 3f 2x 2x

......(i)

Replacing x by x

1 in (i),

2 25f 3f (2x) 2

x x

......(ii)

On solving equation (i) and (ii), we get, 3

8f (2x) 5x 2x

,

5x 6

8f (x) 22 x

2

5 68f (x)

2 x

y xf (x) dy

f (x) xf (x)dx

= 2

1 5x 6 x 5 62

8 2 x 8 2 x

at x 1,dy 1 5 1 5 7

6 2 6dx 8 2 8 2 8

56. (d)

3

3

x 1 , x 1f (x)

1 x , x 1

and

2

2

3x , x 1f (x)

3x , x 1

f (1 ) 3, f (1 ) 3

57. (b)

x 3

2f (x) sin 2x.cos2x.cos3x log 2 ,

2

1f (x) sin 4x cos3x (x 3) log 2

2 ,

1

f (x) [sin 7x sin x] x 34

Differentiate w.r.t. x,

1

f (x) [7cos7x cos x] 14

,

f ( ) 2 1 1 .

58. (b) In neighborhood of 3

x ,4

3 3| cos x | cos x and 3 3| sin x | sin x

3 3y cos x sin x

2 2dy

3cos xsin x 3sin x cos xdx

At 3

x4

,

2 2dy 3 3 3 33cos sin 3sin cos 0

dx 4 4 4 4

.

59. (b)

x

log(log x)f (x) log (log x)

log x

2

1 1log(log x)

x xf (x)(log x)

10

1ef (e)1 e

60. (d)

log x , if 0 x 1

f (x) | log x |log x , if x 1

1, if 0 x 1

xf (x)

1, if x 1

x

.

Clearly f (1 ) 1 and f (1 ) 1 ,

f (x) does not exist at x = 1

61. (c)

Let x xx 1 x 1y log e log e log

x 1 x 1

y x [log(x 1) log(x 1)]

2

dy 1 1 21 1

dx x 1 x 1 (x 1)

2

2

dy x 1

dx x 1

.

62. (a)

1 y x

x exp tanx

1 y xlog x tan

x

y x

tan(log x)x

y x tan(log x) x

2dy sec (log x)

tan(log x) x 1dx x

2dy

tan(log x) sec (log x) 1dx

At x = 1, dy

2dx

.

63. (a)

1 1x 1 x 1

y sec sinx 1 x 1

= 1 1x 1 x 1cos sin

2x 1 x 1

dy

0dx

64. (d)

1 1d cos x sin x dtan tan tan x 1

dx cos x sin x dx 4

.

65. (b)

Let 2 1 1 xy sin cot

1 x

Put 1x cos cos x

2 1 2 11 cosy sin cot sin cot tan

1 cos 2

2y sin2 2

= 2cos

2

=

1(1 cos )

2 =

1(1 x)

2

dy 1

dx 2

66. (a)

Let 5

cos13

. Then 12

sin13

. So, 1y cos {cos .cos x sin .sin x}

1y cos {cos(x )} x ( x is in the first or the second quadrant)

dy

1dx

.

67. (c)

2 2

2 2

2

tan 2x tan x tan 2x tan x tan 2x tan xy cot 3x cot 3x

1 tan 2x tan x 1 tan 2x tan x 1 tan 2x tan x

y tan x tan 3x cot 3x tan x

dysec x

dx

68. (a)

x x

1 x xf (x) cot

2

Put xx tan ,

21 tan 1

y f (x) cot2 tan

= 1cot ( cot 2 ) = 1cot (cot 2 )

y = 2 = 1 x2tan (x )

x

2x

dy 2.x (1 log x)

dx 1 x

f (1) 1 .

69. (a)

2 4 8 16(1 x)(1 x)(1 x )(1 x )(1 x ) 1 x

y1 x 1 x

15 16

2

dy 16x (1 x) 1 x

dx (1 x)

, At x 0 ,

dy1

dx .

70. (c)

2sin x.cos x.cos 2x.cos 4x sin8x

f (x)2sin x 8sin x

2

1 8cos8x.sin x cos x.sin8xf (x) .

8 sin x

f 04

.

71. (a)

x y x y x yxe y 2sin x e xe 1 y' y ' 2cos x

Now x = 0 gives y = 0, hence dy

1dx

.

72. (a)

dy dy 1

sin(3x 2y) log(3x 2y) 3 2 cos 3x 2y 3 2dx dx 3x 2y

dy 3

dx 2

73. (c)

4 5 9 3 5 4 4 8dy dyx y 2(x y) 4x y 5x y 18(x y) 1

dx dx

9 982(x y) 2(x y) dy dy

4 5 18(x y) 1x y dx dx

4 9 9 5 dy

x x y x y y dx

dy y

dx x

74. (b)

dy dy / d

dx dx / d

=a[cos ( sin ) cos ] sin

tana[ sin cos sin ] cos

.

75. (d)

Obviously 1

2

1x cos

1 t

and 1

2

ty sin

1 t

1x tan t and 1y tan t

y x dy

1dx

.

76. (c)

2

2

1 tx

1 t

and

2

2ty

1 t

Put t tan in both the equations to get

2

2

1 tanx cos 2

1 tan

and

2

2 tany sin 2

1 tan

.

Differentiating both the equations, we get dx

2sin 2d

and dy

2cos 2 .d

Therefore dy cos 2 x

dx sin 2 y

.

77. (d)

y x 1 x 1 x 1...to y x 1 y

2y x y 1 dy dy

2y 1dx dx

dy

(2y 1) 1dx

dy 1

dx 2y 1

78. (b)

x 1 ......x 1

y x 1

yy x 1

e elog y ylog x 1

1 dy y dy

ln x 1y dx x 1 dx

1 dy y

ln x 1y dx x 1

2dyx 1 (1 ln y) y

dx

79. (a)

2 2y x

y 2 2y x y 2

2dy dy

2y y.2x xdx dx

2

dy 2xy

dx 2y x

80. (c)

2y xx e

Taking log both sides, log x (2y x)loge 2y x

2y x log x dy 1

2 1dx x

dy 1 x

dx 2x